In my Complex Analysis course the following problem was given after teaching "Zeros of analytic function are isolated" and the Identity Theorem. So I was supposed to solve the problem using above theorems.But I could not manage to solve it using these theorems.Please help me to solve the problem. Thnx in advance.

Prove that $\sin (z+w)= \sin z\cos w +\cos z \sin w \forall z,w \in \mathbb C$

What I done so far:

Fixing $w\in \mathbb R$ and considering $f(z)=\sin(z+w),g(z)=\sin z\cos w +\cos z \sin w$ I get that $f$ and $g$ agree on $\mathbb R$ hence they agree on $\mathbb C$. And this is true $\forall w\in \mathbb R$. But I can not figure out how to solve when $w\in \mathbb C$?

I know the problem can be solved just putting values of $\sin,\cos$ in terms of exponential function but I want to solve using the theorems I have learned.

May be I am missing something very easy,I appologise for that.

Please help me. Thnx again.

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    $\begingroup$ I'm not really sure what the problem is.. The identity theorem says exactly that since the two functions agree on a set with an accumulation point, they must be the same. It doesn't matter what the underlying set really is with regards to $\Bbb C$ (only that it is a subset with an accumulation point). $\endgroup$ Sep 13, 2014 at 13:28
  • $\begingroup$ Thnx for your comment.Yes that is true.But my problem is how to show this identity as it contains two complex variable z and w. If I consider sin(z+w) as a function of z then it agrees with g(z)=sinzcosw +coszsinw in R when w is real.Am I right?then how to handle the case hen w is complex? $\endgroup$
    – usermath
    Sep 13, 2014 at 13:34
  • $\begingroup$ Oh I didn't notice that you had $w$ in the problem statement. That was my bad. I guess your concern is that it is going to lead to a cyclic argument. I'll think about it and make an answer shortly. $\endgroup$ Sep 13, 2014 at 13:39
  • $\begingroup$ Thank you very much Cameron Williams.Waiting for you solution.Thnx again. $\endgroup$
    – usermath
    Sep 13, 2014 at 13:41
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    $\begingroup$ Thank you very much Ian. I am just telling the proof again.Please correct me if I am wrong.So using identity theorem it is true that $\sin (w+x)=\sin w \cos x +\cos w \sin x \forall x \in \mathbb R \forall w\in \mathbb C$ Then I fix $w \in \mathbb C$ to form $f_w(z),g_w(z)$ (as you suugested) and this two agree on $\mathbb R\Rightarrow$ they agree on $\mathbb C$ , And since this is true forall $w\in \mathbb C$ the result follows.Am I right? Thnx again Ian. $\endgroup$
    – usermath
    Sep 13, 2014 at 17:25

1 Answer 1


By fixing $w$ as a real number, you have shown that both functions agree for every $z$ in $C$.

NOW, you can fix $z$ in $C$. Then both the functions agree for every real $w$. This implies (by Identity theorem) they must agree for every $w$ in $C$.


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